If the angle between the circles x^2+y^2-12x- | vedclass

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(B) Given circles: $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$. Centre of the first circle $C_1 = (6, 3)$ and radius $r_1 = \sqrt{6^2+3^2-41} = \sq

Vedclass · Accepted Answer

$-4$

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(B) Given circles: $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$. Centre of the first circle $C_1 = (6, 3)$ and radius $r_1 = \sqrt{6^2+3^2-41} = \sqrt{36+9-41} = \sqrt{4} = 2$. Centre of the second circle $C_2 = (-k/2, -3)$ and radius $r_2 = \sqrt{(-k/2)^2+(-3)^2-(-59)} = \sqrt{k^2/4+9+59} = \sqrt{k^2/4+68}$. Distance between centres $d^2 = (6 - (-k/2))^2 + (3 - (-3))^2 = (6+k/2)^2 + 6^2 = 36 + 6k + k^2/4 + 36 = k^2/4 + 6k + 72$. The angle $	heta$ between two circles is given by $\cos 	heta = \frac{r_1^2+r_2^2-d^2}{2r_1r_2}$. Given $	heta = 45^{\circ}$,so $\cos 45^{\circ} = \frac{1}{\sqrt{2}}$. $\frac{1}{\sqrt{2}} = \frac{4 + (k^2/4 + 68) - (k^2/4 + 6k + 72)}{2(2)\sqrt{k^2/4+68}} = \frac{-6k}{4\sqrt{k^2/4+68}} = \frac{-3k}{2\sqrt{k^2/4+68}}$. Squaring both sides: $\frac{1}{2} = \frac{9k^2}{4(k^2/4+68)} = \frac{9k^2}{k^2+272}$. $k^2+272 = 18k^2$ $\Rightarrow 17k^2 = 272$ $\Rightarrow k^2 = 16$ $\Rightarrow k = \pm 4$. Since $-4$ is an option,the value is $-4$.

If the angle between the circles $x^2+y^2-12x-6y+41=0$ and $x^2+y^2+kx+6y-59=0$ is $45^{\circ}$,then a value of $k$ is

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